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Find the value of a fraction

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Problem:

Find the value of $\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}$.

My Attempt:

For the numerator, I expand them using the Binomial Theorem and get:

$\displaystyle 1^4+2007^4+2008^4$

$\displaystyle=1+(2000+7)^4+(2000+8)^4$

$\displaystyle
=1+2000^4+4(2000)^3(7)^1+6(2000)^2(7)^2+4(2000)^1(7)^3+7^4$

$\displaystyle
+2000^4+4(2000)^3(8)^1+6(2000)^2(8)^2+4(2000)^1(8)^3+8^4$

$\displaystyle
=2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498$


I do the same for the denominator and get:

$\displaystyle 1^2+2007^2+2008^2$

$\displaystyle=1+(2000+7)^2+(2000+8)^2$

$\displaystyle=1+(2000)^2+14(2000)+49+(2000)^2+16(2000)+64$

$\displaystyle=2(2000^2)+30(2000)+114$



I noticed that if I square the denominator, I get twice the value as the numerator, let's see...

$\displaystyle (1^2+2007^2+2008^2)^2$

$\displaystyle
=4(2000^4)+120(2000)^3+1356(2000)^2+6840(2000)+12996$

$\displaystyle
=2(2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498)$


So, we can say that
$\displaystyle \frac{1^4+2007^4+2008^4}{(1^2+2007^2+2008^2)^2}= \frac{1}{2}$

Therefore,
$\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}=\frac{1}{2}(1^2+2007^2+2008^2)=\frac{1}{2} (2(2000^2)+30(2000)+114)=4,030,057$


My question is, are there any other approaches to solve this problem?

Thanks in advance.
 

chisigma

Well-known member
Feb 13, 2012
1,704
You can use the well known relations...

$\displaystyle \sum_{k=1}^{n} k^{2} = \frac{n\ (n+1)}{2}\ \frac{2 n +1}{3}$ (1)

$\displaystyle \sum_{k=1}^{n} k^{4} = \frac{n\ (n+1)}{2}\ \frac{2 n +1}{3}\ \frac{3 n^{2} + 3 n -1}{5} $ (2)

... but I doubt that that is a more comfortable way respect to direct computation...

Kind regards

$\chi$ $\sigma$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
If you know in advance that the fraction is an integer, you can use the following approach.

--------​

Let $f$ be the fraction we want to evaluate. Select $n = 2007$. Then:

$$1^4 + 2007^4 + 2008^4 \equiv 1^4 + 0^4 + 1^4 \equiv 2 \pmod{2007}$$
$$1^2 + 2007^2 + 2008^2 \equiv 1^2 + 0^2 + 1^2 \equiv 2 \pmod{2007}$$
Therefore:

$$f \equiv 1 \pmod{2007} \tag{1}$$
Now use $n = 2008$. We get:

$$1^4 + 2007^4 + 2008^4 \equiv 1^4 + (-1)^4 + 0^4 \equiv 2 \pmod{2008}$$
$$1^2 + 2007^2 + 2008^2 \equiv 1^2 + (-1)^2 + 0^2 \equiv 2 \pmod{2008}$$
And therefore:

$$f \equiv 1 \pmod{2008} \tag{2}$$
And we note that $\gcd{(2007, 2008)} = 1$, and so, using the CRT on (1) and (2):

$$f \equiv 1 \pmod{2007 \times 2008} ~ ~ \implies ~ ~ f \equiv 1 \pmod{4030056} \tag{3}$$
And now, observe that we can write:

$$\frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2} \approx \frac{2 \cdot 2000^4}{2 \cdot 2000^2} \approx 2000^2 \approx 4000000 \tag{4}$$
Which shows that the correct solution must be $f = 4030057$. To be absolutely rigorous, it is enough to prove that:

$$\text{The approximation error is less than the distance between two possible solutions}$$
$$\therefore$$
$$\left | \frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2} - \frac{2 \cdot 2000^4}{2 \cdot 2000^2} \right | < 4030056$$
--------​

This probabilistic approach may be overkill for the problem and is perhaps not what you were looking for, but I thought this was worth posting to illustrate a different -albeit number-theoretical - point of view.
 
Last edited:
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  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
If you know in advance that the fraction is an integer, you can use the following approach.

--------​

Let $f$ be the fraction we want to evaluate. Select $n = 2007$. Then:

$$1^4 + 2007^4 + 2008^4 \equiv 1^4 + 0^4 + 1^4 \equiv 2 \pmod{2007}$$
$$1^2 + 2007^2 + 2008^2 \equiv 1^2 + 0^2 + 1^2 \equiv 2 \pmod{2007}$$
Therefore:

$$f \equiv 1 \pmod{2007} \tag{1}$$
Now use $n = 2008$. We get:

$$1^4 + 2007^4 + 2008^4 \equiv 1^4 + (-1)^4 + 0^4 \equiv 2 \pmod{2008}$$
$$1^2 + 2007^2 + 2008^2 \equiv 1^2 + (-1)^2 + 0^2 \equiv 2 \pmod{2008}$$
And therefore:

$$f \equiv 1 \pmod{2008} \tag{2}$$
And we note that $\gcd{(2007, 2008)} = 1$, and so, using the CRT on (1) and (2):

$$f \equiv 1 \pmod{2007 \times 2008} ~ ~ \implies ~ ~ f \equiv 1 \pmod{4030056} \tag{3}$$
And now, observe that we can write:

$$\frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2} \approx \frac{2 \cdot 2000^4}{2 \cdot 2000^2} \approx 2000^2 \approx 4000000 \tag{4}$$
Which shows that the correct solution must be $f = 4030057$, as the approximation is more than sufficient:

$$\left | \frac{1^4 + 2007^4 + 2008^4}{1^2 + 2007^2 + 2008^2} - \frac{2 \cdot 2000^4}{2 \cdot 2000^2} \right | < 4030056$$
--------​

This probabilistic approach may be overkill for the problem and is perhaps not what you were looking for, but I thought this was worth posting to illustrate a different -albeit number-theoretical - point of view.

Thanks, Bacterius, as you may have already noticed, it's all Greek to me.:eek: But I enjoy reading it very much...(Happy)
 
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  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
You can use the well known relations...

$\displaystyle \sum_{k=1}^{n} k^{2} = \frac{n\ (n+1)}{2}\ \frac{2 n +1}{3}$ (1)

$\displaystyle \sum_{k=1}^{n} k^{4} = \frac{n\ (n+1)}{2}\ \frac{2 n +1}{3}\ \frac{3 n^{2} + 3 n -1}{5} $ (2)

... but I doubt that that is a more comfortable way respect to direct computation...

Kind regards

$\chi$ $\sigma$
Thanks, chisigma...what you've suggested has certainly given me food for thought, but I've got to run now, I will work with it to check if your suggestion could be simplified out neatly and that we don't have to deal with bigger figures too.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Problem:

Find the value of $\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}$.

My Attempt:

For the numerator, I expand them using the Binomial Theorem and get:

$\displaystyle 1^4+2007^4+2008^4$

$\displaystyle=1+(2000+7)^4+(2000+8)^4$

$\displaystyle
=1+2000^4+4(2000)^3(7)^1+6(2000)^2(7)^2+4(2000)^1(7)^3+7^4$

$\displaystyle
+2000^4+4(2000)^3(8)^1+6(2000)^2(8)^2+4(2000)^1(8)^3+8^4$

$\displaystyle
=2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498$


I do the same for the denominator and get:

$\displaystyle 1^2+2007^2+2008^2$

$\displaystyle=1+(2000+7)^2+(2000+8)^2$

$\displaystyle=1+(2000)^2+14(2000)+49+(2000)^2+16(2000)+64$

$\displaystyle=2(2000^2)+30(2000)+114$



I noticed that if I square the denominator, I get twice the value as the numerator, let's see...

$\displaystyle (1^2+2007^2+2008^2)^2$

$\displaystyle
=4(2000^4)+120(2000)^3+1356(2000)^2+6840(2000)+12996$

$\displaystyle
=2(2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498)$


So, we can say that
$\displaystyle \frac{1^4+2007^4+2008^4}{(1^2+2007^2+2008^2)^2}= \frac{1}{2}$

Therefore,
$\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}=\frac{1}{2}(1^2+2007^2+2008^2)=\frac{1}{2} (2(2000^2)+30(2000)+114)=4,030,057$


My question is, are there any other approaches to solve this problem?

Thanks in advance.
I think that the OP's solution is as good as you can get. My approach was to calculate $1+x^4+(x+1)^4 = 2(x^4+2x^3+3x^2 + 2x+1) = 2(x^2+x+1)^2$, and $1+x^2+(x+1)^2 = 2(x^2+x+1)$, so that $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2} = \frac{2(x^2+x+1)^2}{2(x^2+x+1)} = x^2+x+1.$$ That is the same as what anemone did for $x=2007$. Then when you use the fact that $2007 = 2000 + 7$, you easily get the answer $4030057$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think that the OP's solution is as good as you can get. My approach was to calculate $1+x^4+(x+1)^4 = 2(x^4+2x^3+3x^2 + 2x+1) = 2(x^2+x+1)^2$, and $1+x^2+(x+1)^2 = 2(x^2+x+1)$, so that $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2} = \frac{2(x^2+x+1)^2}{2(x^2+x+1)} = x^2+x+1.$$ That is the same as what anemone did for $x=2007$. Then when you use the fact that $2007 = 2000 + 7$, you easily get the answer $4030057$.
I was trying to post this , but the browser just crashed ... :)
 
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  • #8

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
I think that the OP's solution is as good as you can get. My approach was to calculate $1+x^4+(x+1)^4 = 2(x^4+2x^3+3x^2 + 2x+1) = 2(x^2+x+1)^2$, and $1+x^2+(x+1)^2 = 2(x^2+x+1)$, so that $$\frac{1+x^4+(x+1)^4}{1+x^2+(x+1)^2} = \frac{2(x^2+x+1)^2}{2(x^2+x+1)} = x^2+x+1.$$ That is the same as what anemone did for $x=2007$. Then when you use the fact that $2007 = 2000 + 7$, you easily get the answer $4030057$.
Hi Opalg, your approach is much more better than mine because I didn't think of $2(x^4+2x^3+3x^2 + 2x+1)$ could be factorized as $ 2(x^2+x+1)^2$, thanks for giving me a lot of inspirations (both in the past and now) on how to tackle a maths problem more effectively.

Thanks!:)
 

soroban

Well-known member
Feb 2, 2012
409
Hello, anemone

I have a different approach.
I wouldn't claim that it is "better".


$\text{Evaluate: }\:N \;=\;\dfrac{2008^4+2007^4 + 1^4}{2008^2 + 2007^2+1^2}$

Let $u \:=\:2008$

We have: .$N \;=\;\dfrac{u^4 + (u-1)^4 + 1}{u^2 + (u-1)^2 + 1} \;=\;\dfrac{u^4+u^4-4u^3+6u^2-4u+1+1}{u^2+u^2-2u+1+1} $

. . . . . . . . . .$=\; \dfrac{2u^4-4u^3 + 6u^2 - 4u + 2}{2u^2-2u+2} \;=\;\dfrac{2(u^4-2u^3+3u^2 - 2u + 1)}{2(u^2-u+1)} $

. . . . . . . . . .$=\;\dfrac{u^4-2u^3 + 3u^2 - 2u + 1}{u^2-u+1} \;=\;\dfrac{(u^2-u+1)(u^2-u+1)}{u^2-u+1} $

. . . . . . . . . .$=\;u^2-u+1$


Back-substitute: .$N \;=\;2008^2 - 2008 + 1 \;=\;4,\!030,\!057$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

$=\;\dfrac{u^4-2u^3 + 3u^2 - 2u + 1}{u^2-u+1} \;=\;\dfrac{(u^2-u+1)(u^2-u+1)}{u^2-u+1} $

. . . . . . . . . .$=\;u^2-u+1$


Just in case factorizing isn't clear , long division will do the task ...